From linear algebra we know that given vector spaces $V$, $W$ over a field $k$ and a linear map $f\colon V\to W$ we have
$$\dim V = \dim \operatorname{im} f + \dim \ker f.$$
Is this still true when we consider free $\mathbb Z$-modules (i.e. free abelian groups) instead of vector spaces? Given a homomorphism $f\colon G\to H$ between free $\mathbb Z$-modules, do we have
$$\operatorname{rk}(G) = \operatorname{rk}\operatorname{im} f + \operatorname{rk}\ker f?$$
To provide some context: This question comes up when computing homology groups of free chain complexes, where we need to check if some generating set of a kernel is a basis. Using the rank nullity theorem for free $\mathbb Z$-modules is an appealing way to do this.
Best Answer
Here is a proof that doesn't involve going through $\mathbb{Q}$ (and works for any PID):
The image of $f$ is a submodule of a free module, so it is itself free (since $\mathbb{Z}$ is a PID). Therefore the short exact sequence $0 \to \operatorname{ker} f \to G \to \operatorname{im} f \to 0$ is split, and therefore (general fact about split exact sequences of modules) $G \cong \operatorname{ker} f \oplus \operatorname{im} f$. The result about the ranks follows immediately.